lim(x→2) (3ˣ + 3³⁻ˣ - 12) / (3⁻ˣ/² - 3¹⁻ˣ) is equal to

Q: Why can't we just use L'Hôpital's Rule here? It seems faster.

While L'Hôpital's Rule is a powerful tool, differentiating exponential functions with complex powers often leads to messy chain rule applications that increase the probability of calculation errors. Substitution transforms the problem into a polynomial limit, which is much more elegant and less prone to algebraic slips.

Q: How did you know to choose $t = 3^{x/2}$ specifically?

Great question! Look at the exponents: $x$, $3-x$, $-x/2$, and $1-x$. The term $x/2$ is the 'greatest common divisor' of these exponents. By setting $t = 3^{x/2}$, we ensure that all other terms become integer powers of $t$, effectively clearing the exponential clutter.

Q: Is it always safe to cancel the $(t-3)$ term?

Absolutely. In the context of limits, we are analyzing the behavior of the function as $t$ gets arbitrarily close to $3$, but $t$ is never actually equal to $3$. Since $t \neq 3$, the term $(t-3)$ is non-zero, making the cancellation perfectly valid and the only way to resolve the $0/0$ indeterminate form.

Question

$\lim_{x 	o 2} \frac{3^x + 3^{3-x} - 12}{3^{-x/2} - 3^{1-x}}$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a towering, jagged mountain of a problem:

$$\lim_{x 	o 2} \frac{3^x + 3^{3-x} - 12}{3^{-x/2} - 3^{1-x}}$$

At first glance, it looks intimidating. You see exponentials, negative powers, and a fraction that seems designed to confuse.

If you try to plug in $x = 2$ immediately, you are met with the classic $0/0$ indeterminate form. But do not panic; in the world of JEE Advanced, these problems are not walls, but puzzles waiting for the right key.

### Finding the Master Key

The first step in any complex limit problem is to simplify the landscape. We have terms like $3^x$, $3^{3-x}$, $3^{-x/2}$, and $3^{1-x}$. They all share the same base, $3$, but their exponents are a chaotic mix.

We need a common variable to unify them. Look closely at the exponents; the term $x/2$ appears in the denominator. If we set $t = 3^{x/2}$, we can express every other term as a power of $t$.

When we change the variable, we must also change the limit condition. As $x 	o 2$, our new variable $t$ approaches $3^{2/2}$, which is simply $3^1 = 3$. Now, our limit is transformed into $\lim_{t 	o 3}$, a much friendlier territory.

### The Algebraic Dance

Now, let's rewrite the expression. The numerator $3^x$ becomes $(3^{x/2})^2 = t^2$. The term $3^{3-x}$ is $3^3 / 3^x$, which translates to $27 / t^2$.

In the denominator, $3^{-x/2}$ is simply $1/t$, and $3^{1-x}$ becomes $3 / t^2$. Suddenly, the exponential monster has been tamed into a rational function:

$$\lim_{t 	o 3} \frac{t^2 + \frac{27}{t^2} - 12}{\frac{1}{t} - \frac{3}{t^2}}$$

By multiplying the numerator and denominator by $t^2$, we clear the fractions. The numerator becomes $t^4 - 12t^2 + 27$, and the denominator becomes $t - 3$. We have successfully reduced the problem to:

$$\lim_{t 	o 3} \frac{t^4 - 12t^2 + 27}{t - 3}$$

### The Moment of Truth

We are at the final hurdle. We have a polynomial in the numerator and a linear term in the denominator. If we plug in $t = 3$, we still get $0/0$, which tells us that $(t-3)$ is a factor of the numerator.

Let's factorize $t^4 - 12t^2 + 27$. If we treat $t^2$ as a single variable, say $u$, we have $u^2 - 12u + 27$. This factors beautifully into $(u - 9)(u - 3)$, or $(t^2 - 9)(t^2 - 3)$.

Expanding $(t^2 - 9)$ using the difference of squares, we get $(t - 3)(t + 3)(t^2 - 3)$. Now, watch the magic happen as we place this back into our limit:

$$\lim_{t 	o 3} \frac{(t - 3)(t + 3)(t^2 - 3)}{t - 3}$$

The term $(t - 3)$ cancels out perfectly. We are left with $\lim_{t 	o 3} (t + 3)(t^2 - 3)$.

Plugging in $t = 3$, we get $(3 + 3)(3^2 - 3) = 6 	imes 6 = 36$. The mountain has been climbed, and the **final answer is 36**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to 2} \frac{3 ^{x} + 3 ^{3 - x} - 12}{3 ^{- x /2} - 3 ^{1 - x}}$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Limit

The Substitution Strategy

Update the Limit Variable

Express Numerator Terms in $t$

Express Denominator Terms in $t$

Substitute into the Limit

Simplify the Complex Fraction

Cancel Common Denominators

Factorize the Numerator

Apply Difference of Squares

Cancel the Indeterminate Factor

Final Substitution

Final Result

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

Finding the Master Key

The Algebraic Dance

The Moment of Truth

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→2​3−x/2−31−x3x+33−x−12​ is equal to